What does 'p only if q' mean in logic and proofs?

  • Thread starter Thread starter SithsNGiggles
  • Start date Start date
Click For Summary
The discussion clarifies the meaning of "p only if q" in logical implications, specifically in the context of the statement p implies q (p → q). It emphasizes that "p only if q" means that if p is true, then q must also be true; otherwise, the implication is false. Several examples illustrate this concept, demonstrating that the truth of p does not guarantee the truth of q unless q is also true. The conversation acknowledges that while some implications may seem vacuous or irrelevant, they are still logically valid. Understanding this phrasing is essential for grasping logical proofs and truth tables.
SithsNGiggles
Messages
183
Reaction score
0
Interpreting "p implies q"

My Linear Algebra professor had my class work on some proofs, then introduced "truth tables," along with some notation and symbols.

I've taken a class on proofs before, but for some reason it didn't provide any background in pure logic, so I'm a bit lost with one thing my LinAlg prof wrote on the board.

He listed a few ways to interpret
p \Rightarrow q:
  • p implies q,
  • if p then q,
  • q is necessary for p,
  • p is sufficient for q,
  • p only if q

I understand the first four items, but the last one doesn't make sense to me. Can someone please explain how it works? Thanks in advance.
 
Physics news on Phys.org


SithsNGiggles said:
My Linear Algebra professor had my class work on some proofs, then introduced "truth tables," along with some notation and symbols.

I've taken a class on proofs before, but for some reason it didn't provide any background in pure logic, so I'm a bit lost with one thing my LinAlg prof wrote on the board.

He listed a few ways to interpret
p \Rightarrow q:
  • p implies q,
  • if p then q,
  • q is necessary for p,
  • p is sufficient for q,
  • p only if q

I understand the first four items, but the last one doesn't make sense to me. Can someone please explain how it works? Thanks in advance.

I agree that "only if" is the most confusing of the group. I think of it this way.

Say p => q. The only way that can be false is if either p is false, or q is true.

Say p is true. If q is false that makes the implication false. So if p is true then q must be true.

So if p => q is true, then p can be true only if q is true.

Remember, if 2 + 2 = 5 then I am the Pope. That's true.

So 2 + 2 = 5 only if I am the Pope. Can't be any other way.
 


Thanks, SteveL27. The last three lines were very helpful.
 


SithsNGiggles said:
My Linear Algebra professor had my class work on some proofs, then introduced "truth tables," along with some notation and symbols.

I've taken a class on proofs before, but for some reason it didn't provide any background in pure logic, so I'm a bit lost with one thing my LinAlg prof wrote on the board.

He listed a few ways to interpret
p \Rightarrow q:
  • p implies q,
  • if p then q,
  • q is necessary for p,
  • p is sufficient for q,
  • p only if q

I understand the first four items, but the last one doesn't make sense to me. Can someone please explain how it works? Thanks in advance.

The statement is true unless p is true and q is false.

Examples:
"if the moon is green cheese then 2+2=4"

That is true. It seems weird at first, but basically it is saying that 2+2=4 regardless so it doesn't matter what the moon is made of.

"if the moon is green cheese then 2+2=5" Sure. You will never be able to provide a counterexample, so it is a true statement. Vacuous, useless, but true.

"if 1+1=2 then 2+2=4" True. The second statement doesn't follow from the first so it is of no value, but it is indeed true.

'If 1+1=2 then 2+2=5" False!

As you can see, if there is no connection between p and q then any statement relating them is rather vacuous. But there is no harm in that.
 


Thanks ImaLooser. Your examples were pretty helpful too.
 
Greetings, I am studying probability theory [non-measure theory] from a textbook. I stumbled to the topic stating that Cauchy Distribution has no moments. It was not proved, and I tried working it via direct calculation of the improper integral of E[X^n] for the case n=1. Anyhow, I wanted to generalize this without success. I stumbled upon this thread here: https://www.physicsforums.com/threads/how-to-prove-the-cauchy-distribution-has-no-moments.992416/ I really enjoyed the proof...

Similar threads

  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
1
Views
2K